WorksheetsSolving Trig Test Review
Total questions: 43
Worksheet time: 2hrs 58mins
Find the exact value of sin(2x) if sin x = 12/13 and x is in the first quadrant.
cos θ = 4/5 and 270° < θ < 360°
Find sin 2θ
Given cosθ=135 and 23π<θ<2π find cos(2θ)
169120
−169120
2624
2610
10sinxcosx=
sin(10x)
sin(5x)
5sin(2x)
5cos(2x)
Find the exact value of sin 75°
42+6
48
42−6
47
Evaluate cos (90o−x)
cos x
sin x
tan x
−cosx
Find the exact value of tan 12π
Hint: tan(x+y)=cos(x+y)sin(x+y) and tan(x−y)=cos(x−y)sin(x−y)
3+33−3
1+31−3
31
33
Find the exact value of cos 75°
46−2
45+3
21
42−6
Use sum or difference angles identity to find the exact value for sin (−15o)
46+2
46−2
42−6
−23
Use sum or difference angles identity to find the exact value for cos105o
46+2
46−2
4−6−2
42−6
Which of the following is NOT a solution to
sin θ = √(3) / 2 ?
cos2θ = ½
on θ∈[0, 2π)
cosθ = - √(3)/2
on θ∈[0, 2π)
Solve on the interval [0,2π)
tan(x) + 1 = 2
Solve on the interval [0,2π)
2 sinθ + 3 = 2
Solve on the interval [0, 2π)
4sin2x = 3
Solve on the interval [0, 2π)
21sec(x)−1=0
Solve on the interval [0, 2π)
2sin(x)cos(x) = √2 cos(x)
**Hint: get everything on the same side and factor**
Solve on the interval [0, 2π)
cos2(x) + sin(x) + 1 = 0
Solve on the interval [0, 2π)
cos(x) + 2 = 3cos(x)
Which of these is equivalent to
2cos2(x) − 3cos(x) = 0 ?
-cos2x = 0
cosx(2cosx + 3) = 0
cosx(2cosx − 3) = 0
cos x = ⅔
Which is a correct way to solve the equation
cos(x)[2cos(x) − 3] = 0?
divide cos x from both sides
set each factor equal to 0 and solve
distribute cosx into the parentheses
guess and hope for the best
To solve this equation, cos2(x) + sin(x) = 1, replace cos2(x) with
1/(sec2x)
sin2x − 1
1 − sin2x
1 + tan2x
Factor 0 = sin2(x) − sin(x)
0 = sinx(sinx)
0 = sinx(1 − sinx)
0 = cosx(sinx − 1)
0 = sinx(sinx − 1)
True or False?
csc2(x) = 2 is equivalent to sin2(x) = ½
True
False
Factor: sec2(x) − sec(x) − 2
(sec x)(secx − 2)
(secx − 2)(secx − 1)
(secx − 2)(secx + 1)
(secx + 2)(secx − 1)
Solve for x
don't forget cot(x)=sin(x)cos(x)
π
π/3
π/2
-π
Solve: 23secθ + 4 = 0
State all solutions in the interval [0, 2π)
π /3, 2π/3
2π /3, 4π/3
π /6, 5π/6
5π /6, 7π/6
Solve equation for 0≤θ<2π .
2=−4−3cscθ
θ=32π,67π
θ=3π
θ=67π,611π
θ=32π,67π,611π
Solve equation for 0≤θ<2π .
−1−2sec2θ=−3sec2θ
θ=0,π,34π
θ=0
θ=4π,43π,45π,47π
θ=0,π
Solve equation for 0≤θ<2π .
3tan2θ −1 = 0
θ=6π,65π,67π,611π
θ=3π,32π,34π,35π
θ=6π,67π
θ=3π,34π
Solve 2sin2x + sinx − 1 = 0 for 0 ≤ x < 2π
6π, 2π, 65π
2π, 67π, 611π
6π, 65π, 23π
67π, 23π, 611π
Solve cosx tanx + cosx = 0 for 0 ≤ x < 2π
4π, 2π, 45π, 23π
0, 43π, π, 47π
2π, 43π, 23π, 47π
0, 4π, π, 45π
Solve the equation. Restrict your answer to [0,2π).
−22=4sin3θ
{24π,2413π,127π,2425π,2429π,45π,47π}
{127π,2425π,1213π,47π,1223π}
{2425π,2441π,47π}
{125π,127π,1213π,45π,47π,1223π}
State the solutions in the interval [0, 2π)
3tan3x − 3 = 0
18π, 187π, 1813π, 1819π, 1825π, 1831π
18π, 1813π, 1825π
187π, 1819π, 1831π
9π, 94π, 97π, 910π, 913π, 916π
Solve the equation over the interval [0,2π) . Choose all correct answers!
cos(2x)=−21
3π and 35π
32π and 34π
6π and 611π
65π and 67π
43π and 45π
